Transcript pptx

Physics 2102
Gabriela González
Interference
Light is a wave
When two beams of light combine, we can
have constructive or destructive
“interference.”
http://www.colorado.edu/physics/2000/applets/fourier.html
Reflection and refraction
laws
The light travels more slowly in more dense media:
v=c/n (n = index of refraction)
Since the period T is the same, the wavelength has to change (v=/T)

2
sin   v1
sin   , sin  2 



hc
hc
sin  2 2 v2
sin  n2

sin  2 n1
Wavelength:
Frequency:
Snell’s law!
 v1

 2 v2
vn
vn 
n   
c n
c/n c
fn 

  f
n  / n 
Interference: example
A red light beam with wavelength =0.625mm travels through glass (n=1.46) a
distance of 1mm. A second beam, parallel to the first one and originally in phase
with it, travels the same distance through sapphire (n=1.77).
•How many wavelengths are there of each
beam inside the material?
In glass, g=0.625mm/1.46= 0.428 mm and Ng=L/ g=2336.45
In sapphire, s=0.625mm/1.77= 0.353 mm (UV!) and Ns=L/ s=2832.86
•What is the phase difference in the beams when they come out?
The difference in wavelengths is Ns-Ng=496.41.
Each wavelength is 360o, so DN=496.41 means Df=DNx360o=0.41x360o=148o
•How thick should the glass be so that the beams are exactly out of phase at the
exit? (destructive interference!)
DN=L/ s- L/ g= (L/ )(n2-n1)=0.31 (L/ )=m+1/2
A thickness L=(m+0.5) 2.02 mm would make the waves OUT of phase.
For example, 1.009 mm = 499.5 x 2.02 mm makes them come OUT of phase,
and 1.010 mm = 500.0 x 2.02 mm makes them IN phase.
Thin film interference:
The patterns of colors that one sees
in oil slicks on water or in bubbles
is produced by interference of the
light bouncing off the two sides of
the film.
To understand this we
need to discuss the
phase changes that
occur when a wave
moves from one medium
to the another where the
speed is different. This
can be understood with
a mechanical analogy.
Reflection, refraction and changes of phase:
Consider a transverse pulse moving in a rope, that reaches a juncture
with another rope of different density. A reflected pulse is generated.
The reflected pulse is on the same side of
the rope as the incident one if the speed of
propagation in the rope of the right is
faster than on the left.
The reflected pulse is on the opposite side
of the rope if the speed of propagation in the
right is slower than on the left.
The extreme case of no speed on the right corresponds to a rope
anchored to a wall.
If we have a wave instead of a pulse “being on the opposite side of
the rope” means 180 degrees out of phase, or one-half wavelength
out of phase.
Interference: thin films
Phase with respect to incident beam:
r1: f1 = 180o (if n2>n1)
f1 = 0o (if n2<n1)
r2: f2 = 2n2L/ + 180o (if n3>n2)
f1 = 2n2L/
(if n3<n2)
if we have air-oil-water (or air),
Constructive interference → 2n2L/  (2m+1)p
Destructive interference → 2n2L/ = 2m p
Example: mirror coatings
To make mirrors that reflects light of only a given wavelength, a coating of a specific
thickness is used so that there is constructive interference of the given wavelength.
Materials of different index of refraction are used, most commonly MgFe2 (n=1.38) and
CeO2 (n=2.35), and are called “dielectric films”. What thickness is necessary for
reflecting IR light with =1064nm?
First ray: Df=180deg=p
Second ray: Df=2L(2p/(/n))=p
=> L= Ceo2/4(/n)/43nm
Third ray? If wafer has the same
thickness (and is of the same material),
Df=4L(2p/(/n)=2p: destructive!
Choose MgFe2 wafer so that
Df(2n1L1+2n2L2) (2p/)= p+ 2n2L2 (2p/)=3p
> L2= /2n2  386 nm
We can add more layers to keep reflecting the light, until no
light is transmitted: all the light is either absorbed or
reflected.
n=2.35
n=1.38
Huygen’s principle
All points in a wavefront serve as
point sources of spherical
secondary waves.
After a time t, the new
wavefront will be the tangent to all
the resulting spherical waves.
Christian Huygens
1629-1695
Young’s double slit experiment
Young’s double slit experiment
Path difference: DL=d sin 
Bright fringe: DL=m = d sin 
Dark fringe: DL=(m+1/2) = d sin 
The intensity on the screen is
I/I0=4cos2f/2 with f=(2pd/)sin
Michelson interferometers:
As we saw in the previous example, interference is a spectacular way of measuring small
distances (like the thickness of a soap bubble), since we are able to resolve distances of the
order of the wavelength of the light (for instance, for yellow light, we are talking about 0.5
of a millionth of a meter, 500nm). This has therefore technological applications.
In the Michelson interferometer, light from a source (at the
left, in the picture) hits a semi-plated mirror. Half of it goes
through to the right and half goes upwards. The two halves
are bounced back towards the half plated mirror, interfere,
and the interference can be seen by the observer at the
bottom. The observer will see light if the two distances
travelled d1 and d2 are equal, and will see darkness if they
differ by half a wavelength.
Einstein’s messengers (einsteinsmessengers.org)
Michelson-Morley experiment
Michelson won the Nobel prize in 1907, "for his
optical precision instruments and the spectroscopic
and metrological investigations carried out with
their aid"
"The interpretation of these results is that there is no displacement of the
interference bands. ... The result of the hypothesis of a stationary ether is thus
shown to be incorrect." (A. A. Michelson, Am. J. Sci, 122, 120 (1881))
The largest Michelson interferometer in the world is in Livingston, LA,
in LSU owned land (it is operated by a project funded by the National
Science Foundation run by Caltech and MIT, and LSU collaborates in
the project).
http://www.ligo-la.caltech.edu
Mirrors are suspended
with wires and will move
detecting ripples in
the gravitational field due
to astronomical events.
http://www.amnh.org/sciencebulletins/?sid=a.f.gravity.20041101&src=l
American Museum of
Natural History
Science Bulletins
Gravity: Making Waves
Examples
Example
Ocean waves moving at a speed of 4.0m/s are approaching a beach
at an angle of 30o to the normal. The water depth changes abruptly
near the shore, and the wave speed there drops to 3.0m/s. Close to
the beach, what is the direction of wave motion?
Refraction index is (inversely) related to wave speed:
n2/n1=v1/v2=4/3
Snell’s law: n2sin2=n1sin1
2=asin((n1/n2)sin1)
=asin(.75sin(30o))
=22o
Example: Solar panels
Semiconductors such a silicon are used
to build solar cells. They are coated
with a transparent thin film, whose
index of refraction is 1.45, in order to
minimize reflected light. If the index of
refraction of silicon is 3.5, what is the minimum width of the coating
that will produce the least reflection at a wavelength of 552nm?
n=1.45
Both rays undergo 180 phase changes at
reflection, therefore for destructive
interference (no reflection), the distance
travelled (twice the thickness) should be
equal to half a wavelength in the coating
2t 

2n
 95.1nm
Example: the stealth planes
Radar waves have a wavelength of 3cm.
Suppose the plane is made of metal
(speed of propagation=0, n is infinite and
reflection on the polymer-metal surface
therefore has a 180 degree phase change).
The polymer has n=1.5. Same calculation as in previous example
gives,
t

4n

3cm
 0.5cm
4 1.5
On the other hand, if one coated a plane with the same polymer
(for instance to prevent rust) and for safety reasons wanted to maximize
radar visibility (safety!), one would have

3cm
t

 1cm
2n 2 1.5